S4 Physics: Optics Practice Paper 1
- 2026-11-01
S4 Physics: Optics Practice Paper 1
Ten items from reflection and refraction through critical angle, plane mirror images, refractive index and light speed, to convex lens images and magnification.
Quick answer 30 degrees, about 32 degrees, about 48.8 degrees, 4 m, about 1.53, 2 x 10^8 m/s, 15 cm at 0.5x, 30 cm at 2x, n about 1.49, and about 29 degrees.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 Physics: Optics – Tutor Lesson Plan |
| Level | S4 |
02 Questions (10 items)
- FoundationA ray strikes a plane mirror at an angle of incidence of 30 degrees. What is the angle of reflection?
- FoundationLight passes from air into water (refractive index 1.33) at 45 degrees. Find the angle of refraction.
- FoundationLight travels from water to air. Find the critical angle (n = 1.33).
- FoundationAn object stands 2 m in front of a plane mirror. How far apart are the object and its image?
- IntermediateLight enters a medium from air at 50 degrees and refracts at 30 degrees. Find the refractive index.
- IntermediateFind the speed of light in glass of refractive index 1.5.
- IntermediateA convex lens of focal length 10 cm has an object 30 cm away. Find the image distance and magnification.
- ChallengeA convex lens of focal length 10 cm has an object 15 cm away. Find the image distance and magnification.
- ChallengeA medium has a critical angle of 42 degrees. Find its refractive index and the speed of light inside it.
- Error redoA student claims that light entering water at 40 degrees refracts at 40 degrees too. What is wrong, and what is the angle?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 30 degrees | The angle of reflection equals the angle of incidence. |
| 2 | About 32 degrees (sin r = sin 45 / 1.33) | Apply Snell law n = sin i / sin r. |
| 3 | About 48.8 degrees (sin c = 1 / 1.33) | The critical angle comes from sin c = 1 / n. |
| 4 | 4 m (the image is 2 m behind the mirror) | A plane mirror image is as far behind the mirror as the object is in front. |
| 5 | About 1.53 (sin 50 / sin 30) | The refractive index is the ratio of the two sines. |
| 6 | 2 x 10^8 m/s (3 x 10^8 / 1.5) | Since n = c / v, v = c / n. |
| 7 | 15 cm; magnification 0.5, inverted and real | Use 1/f = 1/u + 1/v and m = v / u. |
| 8 | 30 cm; magnification 2, inverted, enlarged and real | With the object between f and 2f the image lies beyond 2f. |
| 9 | n is about 1.49; v is about 2.01 x 10^8 m/s | Find n from sin c = 1 / n, then use v = c / n. |
| 10 | Light bends towards the normal in a denser medium; the angle is about 29 degrees | The two angles are equal only when the angle of incidence is zero. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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